Cayley table

hhmmmm, in what way are the structures of the 3 cayley tables different?

Where have you got the number 3 from? Write out what you've done; I don't think me or DFranklin are following your reasoning at all here.

hhhmmm, since a is the identity element the table will first be

abcd
badc
cdab
dcba

abcd
bcda
cdab
dabc

abcd
bdac
cadb
dcba

This is the only "possibilities" I can think of?

You sure that the first and third are isomorphic? I think it's 2nd and 3rd.

The second and third tables are both cyclic, so they're isomorphic; there are only two non-isomorphic groups of order 4.

ahhh yes, that makes sense but am I correct in saying there are only 3 cayley tables possible?

No; for example:

abcd
badc
cdba
dcab

is another.

Oh yeah. So there are only 4 cayley table possibilties and you was saying there are only 2 non-isomorphic groups, I take it's 1 and 4?

Well 2,3,4 are all isomorphic, i.e. they all represent the same group, but with different elements labelled differently. [That's essentially all an isomorphism is: a relabelling.] So #1 and #2 (which is 'the same' as #3 and #4) are distinct groups. Or you could say #1 and #3 are distinct groups, or #1 and #4 are distinct groups... it doesn't matter.

That's why I thought all of them were isomorphic cause they all was the same but just relabelled which is why I don't understand why 1 isn't isomorphic? all I can think of is that the diagonal starting from the top left going to bottom right all consist of a where as for groups 2,3 and 4 they don't?

In #2,3,4 you have one element which squares to the identity, and two elements which square to that element. That is, your group is of the form $\{ e,x,x^2,x^3 \}$, where $x^2$ is the element that squares to the identity and $x,x^3$ both square to $x^2$. Since you're forced to have $a=e$, it is left open to relabel $x,x^2,x^3$ as $a,b,c$. But since we could let $y=x^3$ and the group would be $\{e,y^3,y^2,y\} = \{e,y,y^2,y^3 \}$ it makes no difference whether an element is x or x³, so the only element where the choice "matters" is $x^2$, hence three different-looking tables (if you put a,b,c,d in that order).

In #1 every element of your group squares to the identity, and so it's of the form $\{ e, x, y, xy \}$. But notice that you could let, say, $p=x, q=xy$, and then you'd have $\{ e, p, pq, q \} = \{ e, p, q, pq \}$, so nothing changes no matter what you call any of the elements, hence only one different-looking table (if you put a,b,c,d in that order).